A079418 -id:A079418 - cp's OEIS frontend

A079416 a(n) = round(prime(n)/n).

2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5

Offset: 1

Reinhard Zumkeller, Jan 07 2003

nonn

The sequence is not monotone, see example and A079417.

			a(20) = round(prime(20)/20) = round(71/20) = round(3.55) = 4;
a(21) = round(prime(21)/21) = round(73/21) = round(3.476190...) = 3;
a(22) = round(prime(22)/22) = round(79/22) = round(3.590909...) = 4.

G. C. Greubel, Table of n, a(n) for n = 1..10000

Cf. A079418, A000040, A038605, A107609, A263076.

[Round(NthPrime(n)/n): n in [1..100]]; // G. C. Greubel, Jan 18 2019

f[n_] := Round[ Prime[n]/n]; Array[f, 105] (* Robert G. Wilson v, Oct 23 2015 *)

vector(100, n, round(prime(n)/n)) \\ G. C. Greubel, Jan 18 2019

[round(nth_prime(n)/n) for n in (1..100)] # G. C. Greubel, Jan 18 2019

A079417 Numbers n such that round(prime(n)/n) < round(prime(n-1)/(n-1)).

Original entry on oeis.org

21, 117, 4117, 4137, 4139, 4142, 4144, 4152, 63326, 63416, 63424, 399872, 399918, 399930, 399944, 399949, 399955, 1014615, 1014635, 1014648, 2582130, 2582200, 2582205, 2582242, 2582374, 2582437, 2582460, 2582483, 2582486, 6592657

Offset: 1

Reinhard Zumkeller, Jan 07 2003

nonn

A079416(a(n)) < A079416(a(n-1)).

Hugo Pfoertner, Table of n, a(n) for n = 1..80

A subsequence of A079418.

Cf. A000040, A079416.

Reap[For[n = 2, n <= 7*10^6, n++, If[Round[Prime[n]/n] < Round[Prime[n-1]/(n-1)], Print[n]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Jun 10 2017 *)

More terms from Sascha Kurz, Jan 09 2003 and from Dean Hickerson, Jan 17 2003

A079419 Primes p such that p/i(p) < prime(i(p)-1)/(i(p)-1), where i(p) = A049084(p).

Original entry on oeis.org

3, 13, 19, 31, 43, 61, 73, 103, 109, 131, 139, 151, 167, 181, 193, 197, 199, 227, 229, 233, 241, 271, 281, 283, 311, 313, 317, 349, 353, 383, 401, 421, 433, 443, 461, 463, 467, 491, 503, 523, 571, 601, 617, 619, 643, 647, 661, 677, 743, 761, 773, 811, 823

Offset: 1

Reinhard Zumkeller, Jan 07 2003

nonn

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000040, A049084, A079418.

seq[max_] := Module[{p = Prime[Range[max]]}, p[[1 + Position[Differences[p/Range[max]], ?(# < 0 &)] // Flatten]]]; seq[150] (* _Amiram Eldar, Mar 25 2025 *)

a(n) = A000040(A079418(n)).

A342877 a(n) = 1 if the average distance between consecutive first n primes is greater than that of the first n-1 primes, otherwise a(n) = 0, for n > 2.

Original entry on oeis.org

1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1

Offset: 3

Andres Cicuttin, Mar 28 2021

nonn

The average distance between consecutive primes among the first n primes tends to increase with n. This average distance always changes when n is increased to n + 1, but it seems that most of the times this distance decreases. See a log-linear scatter plot of (1/n) Sum_{i=1..n} a(i) in Links.
Conjecture: lim_{n->infinity} (1/n) Sum_{i=1..n} a(i) < 1/2.
In support of the conjecture: If it is assumed, as an approximation, that position of primes follows a Poisson point process then the distance between consecutive primes is a stochastic variable with exponential probability distribution function. The probability that an exponentially distributed stochastic variable takes a value larger than the mean value is about 0.367879.

			a(3) = 1 because the average distance between consecutive first three primes {2,3,5} is (5 - 2)/2 = 3/2 which is greater than the average distance between consecutive first two primes {2,3} which is (3-2)/1 = 1.
a(6)=0 because the average distance between consecutive first six primes {2,3,5,7,11,13} is (13 - 2)/5 = 11/5 which is smaller than the average distance between consecutive first five primes {2,3,5,7,11} which is (11 - 2)/4 = 9/4.

Andres Cicuttin, Log-linear scatter plot of (1/n) Sum_{i=1..n} a(i) for first 2^18 primes.
Wikipedia, Poisson point process
Wikipedia, Exponential distribution
Yamasaki, Yasuo and Yamasaki, Aiichi, On the Gap Distribution of Prime Numbers, 数理解析研究所講究録 (1994), 887: 151-168.
Index entries for characteristic functions

Cf. A001223, A079418, A286888.

a={}; nmax=128;
Do[If[(Prime[n]-2)/(n-1)>(Prime[n-1]-2)/(n-2),AppendTo[a,1],AppendTo[a,0]],{n,3,nmax}];
a
(* Uncomment and run next lines to produce the log-linear plot available in Links *)
(* a={};
nmax=2^18;
Do[If[(Prime[n]-2)/(n-1)>(Prime[n-1]-2)/(n-2),AppendTo[a,{n,1}],AppendTo[a,{n,0}]],{n,3,nmax}];
ListLogLinearPlot[Transpose[{Range[3,nmax],Accumulate[Transpose[a][[2]]]/Range[3,nmax]}],Frame->True,PlotRange->{All,{0.25,0.75}},PlotLabel->Text[Style["Sum_{i=1..n} a(i)/n",FontSize->16]],
FrameLabel->{Text[Style["n",FontSize->16]],},PlotStyle->{PointSize->Small,Red},GridLines->Automatic] *)

A342877(n) = (((prime(n)-2)/(n-1)) > ((prime(n-1)-2)/(n-2))); \\ Antti Karttunen, Mar 28 2021

cp's OEIS Frontend

A079416 a(n) = round(prime(n)/n).

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A079417 Numbers n such that round(prime(n)/n) < round(prime(n-1)/(n-1)).

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A079419 Primes p such that p/i(p) < prime(i(p)-1)/(i(p)-1), where i(p) = A049084(p).

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A342877 a(n) = 1 if the average distance between consecutive first n primes is greater than that of the first n-1 primes, otherwise a(n) = 0, for n > 2.

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